Standard Methods for Examination of Water & Wastewater_11 potx

Coagulation

Colloids are agglomerates of atoms or molecules whose sizes are so small that
gravity has no effect on settling them but, instead, they stay in suspension. Because
they stay in suspension, they are said to be stable. The reason for this stability is
the mutual repulsion between colloid particles. They may, however, be destabilized
by application of chemicals.

Coagulation

is the unit process of applying these
chemicals for the purpose of destabilizing the mutual repulsion of the particles, thus
causing the particles to bind together. This process is normally applied in conjunction
with the unit operation of flocculation. The colloid particles are the cause of the
turbidity and color that make waters objectionable, thus, should, at least, be partially
removed.
This chapter applies the techniques of the unit process of coagulation to the
treatment of water and wastewater for the removal of colloids that cause turbidity
and color. It also discusses prerequisite topics necessary for the understanding of
coagulation such as the behavior of colloids, zeta potential, and colloid stability. It
then treats the coagulation process, in general, and the unit process of the use of
alum and the iron salts, in particular. It also discusses chemical requirements and
sludge production.

12.1 COLLOID BEHAVIOR

Much of the suspended matter in natural waters is composed of silica, or similar
materials, with specific gravity of 2.65. In sizes of 0.1 to 2 mm, they settle rapidly;
however, in the range of the order of 10

−

5

mm, it takes them a year, in the overall,
to settle a distance of only 1 mm. And, yet, it is the particle of this size range that
causes the turbidity and color of water, making the water objectionable. The removal
of particles by settling is practical only if they settle rapidly in the order of several
hundreds of millimeters per hour. This is where coagulation can perform its function,
by destabilizing the mutual repulsions of colloidal particles causing them to bind
together and grow in size for effective settling. Colloidal particles fall in the size
range of 10

−

6

mm to 10

−

3

mm. They are aggregates of several hundreds of atoms or
molecules, although a single molecule such as those of proteins is enough to be
become a colloid. The term colloid comes from the two Greek words

kolla

, meaning

glue, and

eidos

, meaning like.
A colloid system is composed of two phases: the

dispersed phase

, or the

solute

,
and the

dispersion medium

, or the

solvent.

Both of these phases can have all three
states of matter which are solid, liquid, and gas. For example, the dispersion medium
may be a liquid and the dispersed phase may be a solid. This system is called a

liquid
sol

, an example of which is the turbidity in water. The dispersion medium may be a

12

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546

gas and dispersed phase may be solid. This system is called a

gaseous sol

, and examples
are dust and smoke. Table 12.1 shows the different types of colloidal systems. Note
that it is not possible to have a colloidal system of gas in a gas, because gases are
completely soluble in each other. In the coagulation treatment of water and wastewater,
we will be mainly interested in the solid being dispersed in water, the liquid sol or
simply sol. Unless required for clarity, we will use the word ‘‘sol’’ to mean liquid sol.
Sols are either

lyophilic

or

lyophobic

. Lyophilic sols are those that bind the solvent,
while the lyophobic sols are those that do not bind the solvent. When the solvent is water,
lyophilic and lyophobic sols are, respectively, called

hydrophilic

and

hydrophobic sols

.
The affinity of the

hydrophilic sols

for water is due to polar functional groups that
exist on their surfaces. These groups include such polar groups as

−

OH,

−

COOH, and

−

NH

2

. They are, respectively, called the

hydroxyl, carboxylic

, and

amine

groups.
Figure 12.1a shows the schematic of a hydrophilic colloid. As portrayed, the functional
polar groups are shown sticking out from the surface of the particle. Because of the
affinity of these groups for water, the water is held tight on the surface. This water is
called

bound water

and is fixed on the surface and moves with the particle.
The

hydrophobic colloids

do not have affinity for water; thus, they do not contain
any bound water. In general, inorganic colloids are hydrophobic, while organic
colloids are hydrophilic. An example of an inorganic colloid is the clay particles
that cause turbidity in natural water, and an example of an organic colloid is the
colloidal particles in domestic sewage.

12.2 ZETA POTENTIAL

The repulsive property of colloid particles is due to electrical forces that they possess.
The characteristic of these forces is indicated in the upper half of Figure 12.1b. At
a short distance from the surface of the particle, the force is very high. It dwindles
down to zero at infinite distance from the surface.

TABLE 12.1
Types of Colloidal Systems

Dispersion
Medium Dispersed Phase Common Name Example

Solid Solid Solid sol Colored glass and gems, some alloys
Solid Liquid Solid emulsion Jelly, gel, opal (SiO

2

and H

2

O), pearl
(CaCO

3

and H

2

O)
Solid Gas None Pumice, floating soap
Liquid Solid Liquid sol Turbidity in water, starch suspension,
ink, paint, milk of magnesia
Liquid Liquid Liquid emulsion Oil in water, milk, mayonnaise, butter

Liquid Gas Foam Whipped cream, beaten egg whites
Gas Solid Gaseous sol Dust, smoke
Gas Liquid Gaseous emulsion Mist, fog, cloud, spray
Gas Gas Not applicable None

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547

The electrical forces are produced due to the charges that the particles possess
at their surfaces. These charges called

primary charges

are, in turn, produced from
one or both of two phenomena: the dissociation of the polar groups and preferential
adsorption of ions from the dispersion medium. The primary charges on hydrophobic
colloids are due to preferential adsorption of ions from the dispersion medium.
The primary charges on hydrophilic colloids are due chiefly to the polar groups
such as the carboxylic and amine groups. The process by which the charges on these
types of colloids are produced is indicated in Figure 12.2. The symbol

R

represents
the colloid body. First, the colloid is represented at the top of the drawing, without the
effect of pH. Then by a proper combination of the H

+

and OH

−

being added to the
solution, the colloid attains ionization of both carboxylic and the amine groups. At
this point, both ionized groups neutralize each other and the particle is neutral. This
point is called the

isoelectric point

, and the corresponding ion of the colloid is called
the

zwitter ion

. Increasing the pH by adding a base cause the added OH

−

to neutralize
the acid end of the zwitter ion (the ); the zwitter ion disappears, and the whole
particle becomes negatively charged. The reverse is true when the pH is reduced by
the addition of an acid. The added H

+

neutralizes the base end of the zwitter ion
(the COO

−

); the zwitter ion disappears, and the whole particle becomes positively
charged. From this discussion, a hydrophilic colloid can attain a primary charge of
either negative or positive depending upon the pH.
The primary charges on a colloid which, as we have seen, could either be positive
or negative, attract ions of opposite charges from the solution. These opposite charges
are called

counterions

. This is indicated in Figure 12.3. If the primary charges are

FIGURE 12.1

(a) Hydrophilic colloid encased in bound water; (b) interparticle forces as a
function of interparticle distance.
(a) (b)
Bound water
Attraction
Attraction
Force
Repulsion
Repulsion
Resultant
Distance
COOH
COOH
COOH

HOOC
HOOC
NH
2
NH
2
NH
2
H
2
N
H
2
N
Colloid
body
NH
3
+

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548

sufficiently large, the attracted counterions can form a compact layer around the
primary charges. This layer is called the

Stern layer

. The counterions, in turn, can
attract their own counterions, the

coions

of the primary charges, forming another
layer. Since these coions form a continuous distribution of ions into the bulk of the
solution, they tend to be diffused and form a diffused layer. The second layer is
called the

Gouy layer

. Thus, the

Stern

and

Gouy

layers form an envelope of electric
double layer around the primary charges.
All of the charges in the Stern layer move with the colloid; thus, this layer is a
fixed layer. In the Gouy layer, part of the layer may move with the colloid particle
by shearing at a

shear plane

. This layer may shear off beyond the boundary of the
fixed Stern layer measured from the surface of the colloid. Thus, some of the charges

in the layer move with the particle, while others do not. This plane is indicated in
Figure 12.3.
The charges are electric, so they possess electrostatic potential. As indicated on
the right-hand side of Figure 12.3, this potential is greatest at the surface and decreases
to zero at the bulk of the solution. The potential at a distance from the surface at
the location of the shear plane is called the

zeta potential

. Zeta potential meters are
calibrated to read the value of this potential. The greater this potential, the greater
is the force of repulsion and the more stable the colloid.

12.3 COLLOID DESTABILIZATION

Colloid stability may further be investigated by the use Figure 12.1b. This figure
portrays the competition between two forces at the surface of the colloid particle: the

van der Waal’s force of attraction

, represented by the lower dashed curve, and the
force of repulsion, represented by the upper dashed curve. The solid curve represents
the resultant of these two forces. As shown, this resultant becomes zero at

a

−

a

′

and
becomes fully an attractive force to the left of the line. When the resultant force
becomes fully attractive, two colloid particles can bind themselves together.

FIGURE 12.2

Primary charges of a hydrophilic colloid as a function of pH.
COOH
COOH
NH
2
NH
3
+
NH
3
+
NH
3
OH
COO
–
OH
–
COO
–
R
pH

RR
R
H
+
H,
+
OH
–
Isoelectric point

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549
FIGURE 12.3

Charged double layer around a negatively charged colloid particle (left) and variation of electrostatic potential with distance from particle
surface (right).
Bulk of solution
Diffuse layer
Fixed layer
Electronegative particle
Shear
plane
Electrostatic potential
Plane of shear
Zeta
potential
Fixed
layer

Diffuse
layer
Solution
bulk
Distance from particle surface

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550

Physical–Chemical Treatment of Water and Wastewater

The force of repulsion, as we have seen, is due to the charges on the surface.
Inherent in any body is a natural force that tends to bind particles together. This
force is exactly the same as the force that causes adsorption of particles to an
adsorbing surface. This is caused by the imbalance of atomic forces on the surface.
Whereas atoms below the surface of a particle are balanced with respect to forces
of neighboring atoms, those at the surface are not. Thus, the unbalanced force at the
surface becomes the van der Waal’s force of attraction. By the presence of the primary
charges that exert the repulsive force, however, the van der Waal’s force of attraction
is nullified until a certain distance designated by

a

−



a

′

is reached. The distance can
be shortened by destabilizing the colloid particle.
The use of chemicals to reduce the distance to

a

−

a

′

from the surface of the
colloid is portrayed in Figure 12.4. The zeta potential is the measure of the stability
of colloids. To destabilize a colloid, its zeta potential must be reduced; this reduction
is equivalent to the shortening of the distance to

a

−

a

′

and can be accomplished
through the addition of chemicals.

The chemicals to be added should be the counterions of the primary charges.
Upon addition, these counterions will neutralize the primary charges reducing the
zeta potential. This process of reduction is indicated in Figures 12.4a and 12.4b; the
potential is reduced in going from Figure 12.4a to 12.4b. Note that destabilization
is simply the neutralization of the primary charges, thus reducing the force of
repulsion between particles. The process is not yet

the coagulation

of the colloid.

FIGURE 12.4

Reduction of zeta potential to cause destabilization of colloids.
Fixed
layer
Fixed
layer
Diffuse
layer
Diffuse
layer
Zeta
potential
Zeta
potential
Electrostatic potential
Plane of shear
Distance from particle surface
(a) Prior to addition of counter ions

Distance from particle surface
(b) After addition of counter ions

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551

12.4 COAGULATION PROCESS

The destabilization of colloids through the addition of counterions should be done
in conjunction with the application of the complete coagulation process. Four meth-
ods are used to bring about this process: double-layer compression, charge neutral-
ization, entrapment in a precipitate, and intraparticle bridging.
When the concentration of counterions in the dispersion medium is smaller, the
thickness of the electric double layer is larger. Two approaching colloid particles
cannot come close to each other because of the thicker electric double layer, there-
fore, the colloid is stable. Now, visualize adding more counterions. When the con-
centration is increased, the attracting force between the primary charges and the
added counterions increases causing the double layer to shrink. The layer is then
said to be compressed. As the layer is compressed sufficiently by the continued
addition of more counterions, a time will come when the van der Waals force exceeds
the force of repulsion and coagulation results.
The charge of a colloid can also be directly neutralized by the addition of ions
of opposite charges that have the ability to directly adsorb to the colloid surface.
For example, the positively charged dodecylammoniun, C

12

H

25

, tends to be
hydrophobic and, as such, penetrates directly to the colloid surface and neutralize
it. This is said to be a direct charge neutralization, since the counterion has penetrated
directly into the primary charges. Another direct charge neutralization method would
be the use of a colloid of opposite charge. Direct charge neutralization and the
compression of the double layer may compliment each other.
A characteristic of some cations of metal salts such as Al(III) and Fe(III) is that
of forming a precipitate when added to water. For this precipitation to occur, a
colloidal particle may provide as the seed for a nucleation site, thus, entrapping the
colloid as the precipitate forms. Moreover, if several of this particles are entrapped
and are close to each other, coagulation can result by direct binding because of the
proximity.
The last method of coagulation is intraparticle bridging. A bridging molecule
may attach a colloid particle to one active site and a second colloid particle to another
site. An

active site

is a point in the molecule where particles may attach either by
chemical bonding or by mere physical attachment. If the two sites are close to each
other, coagulation of the colloids may occur; or, the kinetic movement may loop the
bridge assembly around causing the attached colloids to bind because for now they
are hitting each other, thus bringing out coagulation.

12.4.1 C

OAGULANTS

FOR



THE

C

OAGULATION

P

ROCESS



Electrolytes and polyelectrolytes are used to coagulate colloids.

Electrolytes

are
materials which when placed in solution cause the solution to be conductive to
electricity because of charges they possess.

Polyelectrolytes

are polymers possessing

more than one electrolytic site in the molecule, and

polymers

are molecules joined
together to form larger molecules. Because of the charges, electrolytes and poly-
electrolytes coagulate and precipitate colloids. The coagulating power of electrolytes
is summed up in the

Schulze–Hardy rule

that states: the coagulation of a colloid
is affected by that ion of an added electrolyte that has a charge opposite in sign to
NH
3
+

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552

that of the colloidal particle; the effect of such an ion increases markedly with the
number of charges carried. Thus, comparing the effect of AlCl

3

and Al

2

(SO

4

)

3

in
coagulating positive colloids, the latter is 30 times more effective than the former,
since sulfate has two negative charges while the chloride has only one. In coagulating
negative colloids, however, the two have about the same power of coagulation.
The most important coagulants used in water and wastewater treatment are alum,
copperas (ferrous sulfate), ferric sulfate, and ferric chloride. Later, we will specifi-
cally discuss the chemical reactions of these coagulants at greater lengths. Other
coagulants have also been used but, owing to high cost, their use is restricted only
to small installations. Examples of these are sodium aluminate, NaAlO

2

; ammonia
alum, ; and potash alum,
The reactions of sodium aluminate with aluminum sulfate and carbon dioxide are:
(12.1)
(12.2)

12.4.2 C

OAGULANT

A

IDS

Difficulties with settling often occur because of flocs that are slow-settling and are
easily fragmented by the hydraulic shear in the settling basin. For these reasons,
coagulant aids are normally used. Acids and alkalis are used to adjust the pH to the
optimum range. Typical acids used to lower the pH are sulfuric and phosphoric acids.
Typical alkalis used to raise the pH are lime and soda ash. Polyelectrolytes are also
used as coagulant aids. The cationic form has been used successfully in some waters
not only as a coagulant aid but also as the primary coagulant. In comparison with
alum sludges that are gelatinous and voluminous, sludges produced by using cationic
polyelectrolytes are dense and easy to dewater for subsequent treatment and disposal.
Anionic and nonionic polyelectrolytes are often used with primary metal coagulants
to provide the particle bridging for effective coagulation. Generally, the use of poly-
electrolyte coagulant aids produces tougher and good settling flocs.
Activated silica and clays have also been used as coagulant aids. Activated silica
is sodium silicate that has been treated with sulfuric acid, aluminum sulfate, carbon
dioxide, or chlorine. When the activated silica is applied, a stable negative sol is
produced. This sol unites with the positively charged primary-metal coagulant to
produce tougher, denser, and faster settling flocs.
Bentonite clays have been used as coagulant aids in conjunction with iron and
alum primary coagulants in treating waters containing high color, low turbidity, and
low mineral content. Low turbidity waters are often hard to coagulate. Bentonite
clay serves as a weighting agent that improves the settleability of the resulting flocs.

12.4.3 R

APID

M

IX



FOR

C

OMPLETE

C

OAGULATION

Coagulation will not be as efficient if the chemicals are not dispersed rapidly
throughout the mixing tank. This process of rapidly mixing the coagulant in the
volume of the tank is called

rapid

or

flash mix

. Rapid mixing distributes the chemicals
immediately throughout the volume of the mixing tank. Also, coagulation should
Al

2
(SO
4
)
3
(NH
4
)
2
24H
2
O⋅⋅
Al
2
(SO
4
)
3
K
2
SO
4
24H
2
O.⋅⋅
6NaAlO
2
Al
2
+ SO

4
()
3
14.3H
2
O 8Al OH()
3
3Na
2
SO
4
2.3H
2
O++→⋅
2NaAlO
2
CO
2
3H
2
O++ 2Al OH()
3
↓
Na
2
CO
3
+→
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be followed by flocculation to agglomerate the tiny particles formed from the coag-
ulation process.
If the coagulant reaction is simply allowed to take place in one portion of the
tank because of the absence of the rapid mix rather than being spread throughout
the volume, all four mechanisms for a complete coagulation discussed above will not
be utilized. For example, charge neutralization will not be utilized in all portions of
the tank because, by the time the coagulant arrives at the point in question, the reaction
of charge neutralization will already have taken place somewhere.
Interparticle bridging will not be as effective, since the force to loop the bridge
around will not be as strong without the force of the rapid mix. Colloid particles
will not effectively be utilized as seeds for nucleation sites because, without rapid
mix, the coagulant may simply stay in one place. Finally, the compression of the
double layer will not be as effective if unaided by the force due to the rapid mix.
The force of the rapid mix helps push two colloids toward each other, thus enhancing
coagulation. Hence, because of all these stated reasons, coagulation should take
place in a rapidly mixed tank.
12.4.4 THE JAR TEST
In practice, irrespective of what coagulant or coagulant aid is used, the optimum
dose and pH are determined by a jar test. This consists of four to six beakers (such
as 1000 ml in volume) filled with the raw water into which varying amounts of dose
are administered. Each beaker is provided with a variable-speed stirrer capable of
operating from 0 to 100 rpm.
Upon introduction of the dose, the contents are rapidly mixed at a speed of about
60 to 80 rpm for a period of one minute and then allowed to flocculate at a speed
of 30 rpm for a period of 15 minutes. After the stirring is stopped, the nature and
settling characteristics of the flocs are observed and recorded qualitatively as poor,
fair, good, or excellent. A hazy sample denotes poor coagulation; a properly coag-
ulated sample is manifested by well-formed flocs that settle rapidly with clear water
between flocs. The lowest dose of chemicals and pH that produce the desired flocs
and clarity represents the optimum. This optimum is then used as the dose in the

actual operation of the plant. See Figure 12.5 for a picture of a jar testing apparatus.
12.5 CHEMICAL REACTIONS OF ALUM
The alum used in water and wastewater treatment is Al
2
(SO
4
)
3
⋅ 14H
2
O. (The ‘‘14’’
actually varies from 13 to 18.) For brevity, this will simply be written without the
water of hydration as Al
2
(SO
4
)
3
. When alum is dissolved in water, it dissociates
according to the following equation (Sincero, 1968):
(12.3)
By rapid mix, the ions must be rapidly dispersed throughout the tank in order to
effect the complete coagulation process.
Al
2
(SO
4
)
3
2Al

3+
→ 3SO
4
2−
+
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Because the water molecule is polar, it attracts Al
3+
forming a complex ion
according to the following:
(12.4)
In the complex ion , Al is called the central atom and the molecules of
H
2
O are called ligands. The subscript 6 is the coordination number, the number of
ligands attached to the central atom; the superscript 3+ is the charge of the complex
ion. The whole assembly of the complex forms what is called a coordination sphere.
As indicated in Equation (12.4), aluminum has a coordination number of 6 with
the water molecule. This means that no more water molecules can bind with the central
atom but that any interaction would not be a mere insertion into the coordination
sphere. In fact, further reaction with the water molecule involves hydrolysis of the
water molecule and exchanging of the resulting OH
−
ion with the H
2
O ligand inside
the coordination sphere. This type of reaction is called ligand exchange reaction.
Some of the hydrolysis products of the ligand exchange reaction are mononu-
clear, which means that only one central atom of aluminum is in the complex; and

some are polynuclear, which means that more than one central atom of aluminum
exists in the complex. Because the water molecule is not charged, may
simply be written as Al
3+
. This is the symbol to be used in the complex reactions
that follow. Without going into details, we will simply write at once all the complex
ligand exchange equilibrium reactions.
(12.5)
(12.6)
(12.7)
FIGURE 12.5 A Phipps and Bird jar testing apparatus. (Courtesy of Phipps & Bird,
Richmond, VA. © 2002 Phipps & Bird.)
Al
3+
6H
2
O+ Al(H
2
O)
6
3+
→
Al(H
2
O)
6
3+
Al(H
2
O)

6
3+
Al
3+
H
2
O  Al(OH)
2+
H
+
++ K
Al(OH)c
10
−5
=
7Al
3+
17H
2
O  Al
7
(OH)
17
4+
17H
+
++
K
Al
7

(OH)
17
c
10
−48.8
=
13Al
3+
34H
2
O  Al
13
(OH)
34
5+
34H
+
++
K
Al
13
(OH)
34
c
10
−97.4
=
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© 2003 by A. P. Sincero and G. A. Sincero
(12.8)

(12.9)
(12.10)
The equilibrium constants apply at 25°C:
Note: From the number of aluminum atoms they contain , ,
and are polynuclear complexes.
Also, the H
+
and the OH
−
are participants in these reactions. This means that the
concentrations of each of these complex ions are determined by the pH of the solution.
In the application of the previous equations in coagulation treatment of water,
conditions must be adjusted to allow maximum precipitation of the solid represented
by To allow for this maximum precipitation, the concentrations of the
complex ions must be held to a minimum.
12.5.1 DETERMINATION OF THE OPTIMUM pH
For effective removal of the colloids, as much of alum should be converted to the
solid . Also, as much of the concentrations of the complex ions should
neutralize the primary charges of the colloids to effect their destabilization. Overall,
this means that once the solids have been formed and the complex ions have
neutralized the colloid charges, the concentrations of the complex ions standing in
solution should be at the minimum. The pH corresponding to this condition is called
the optimum pH.
Let sp
Al
represent all the species that contain the aluminum atom standing in solution.
Thus, the concentration of all the species containing the aluminum atom Al(III), is
(12.11)
All the concentrations in the right-hand side of the previous equation will now be
expressed in terms of the hydrogen ion concentration. This will result in the express-

ing of [sp
Al
] in terms of the hydrogen ion. Differentiating the resulting equation of
[sp
Al
] with respect to [H
+
] and equating the result to zero will produce the minimum
concentration of sp
Al
and, thus, the optimum pH determined. Using the equilibrium
reactions, Eqs. (12.5) through (12.10), along with the ion product of water, we now
proceed as follows:
(12.12)
Al(OH)
3 s()
(fresh precipitate)  Al
3+
3OH
−
+ K
sp,Al(OH)
3
10
−33
=
Al(OH)
3 s()
OH
−

 Al OH()
4
−
+ K
Al(OH)
4
c
10
+1.3
=
2Al
3+
2H
2
O  Al
2
OH()
2
4+
2H
+
++
K
Al
2
(OH)
2
c
10
−6.3

=
Al
7
(OH)
17
4+
Al
13
(OH)
34
5+
Al
2
(OH)
2
4
+
Al(OH)
3 s()
.
Al(OH)
3 s()
sp
Al
[] Al
3+
[]Al OH()
2+
[]7Al
7

OH()
17
4+
[]13 Al
13
OH()
34
5+
[]++ +=
Al OH()
4
−
[]2Al
2
OH()
2
4+
[]++
Al
3+
[]
Al
3+
{}
γ
Al

K
sp,Al(OH)
3

γ
Al
{OH
−
}
3

K
sp,Al(OH)
3
H
+
{}
3
γ
Al
K
w
3

K
sp,Al(OH)
3
γ
H
3
H
+
[]
3

γ
Al
K
w
3

== = =
TX249_frame_C12.fm Page 555 Friday, June 14, 2002 2:29 PM
© 2003 by A. P. Sincero and G. A. Sincero
(12.13)
(12.14)
(12.15)
(12.16)
(12.17)
, , are, respectively, the
activity coefficients of the aluminum ion and the hydrogen ion and the complexes
Al(OH)
2+
, , , , and . is the
solubility product constant of the solid and K
w
is the ion product of
water. , , , and are, respectively, the
equilibrium constants of the complexes , , ,
and .
Remember that the equilibrium constants are a function of temperature. To obtain
the corresponding values at other temperatures, the Van’t Hoff equation should be
used. The use of this equation, however, requires the value of the standard enthalpy
. At present, none are available for the aluminum complexes. Research is
therefore needed to find these values.

Al OH()
2+
[]
Al OH()
2+
{}
γ
Al OH()c

K
Al OH()c
Al
3+
{}
γ
Al OH()c
H
+
{}

K
Al OH()c
K
sp,Al OH()
3
γ
H
2
H
+

[]
2
γ
Al OH()c
K
w
3

== =
Al
7
(OH)
17
4+
[]
Al
7
(OH)
17
[]
γ
Al
7
(OH)
17
c

K
Al
7

(OH)
17
c
Al
3+
{}
7
γ
Al
7
(OH)
17
c
H
+
{}
17

K
Al
7
(OH)
17
c
γ
Al
7
Al
3+
[]

7
γ
Al
7
(OH)
17
c
γ
H
17
H
+
[]
17

== =
K
Al
7
(OH)
17
c
K
sp,Al(OH)
3
7
γ
H
4
[H

+
]
4
γ
Al
7
(OH)
17
c
K
w
21

=
[Al
13
(OH)
34
5+
]
Al
13
OH()
34
5+
{}
γ
Al
13
(OH)

34
c

K
Al
13
(OH)
34
c
Al
3+
{}
13
γ
Al
13
(OH)
34
c
H
+
{}
34

K
Al
13
(OH)
34
c

γ
Al
13
Al
3+
[]
13
γ
Al
13
(OH)
34
c
γ
H
34
H
+
[]
34

== =
K
Al
13
(OH)
34
c
K
sp,Al(OH)

3
13
γ
H
5
H
+
[]
5
γ
Al
13
(OH)
34
c
K
w
39

=
[Al(OH)
4
−
]
{Al(OH)
4
−
}
γ
Al(OH)

4
c

K
Al(OH)
4
c
{OH
−
}
γ
Al(OH)
4
c

K
Al(OH)
4
c
K
w
γ
Al(OH)
4
c
γ
H
[H
+
]

== =
[Al
2
(OH)
2
4+
]
{Al
2
(OH)
2
4+
}
γ
Al
2
(OH)
2
c

K
Al
2
(OH)
2
c
{Al
3+
}

2
γ
Al
2
(OH)
2
c
{H
+
}
2

K
Al
2
(OH)
2
c
γ
Al
2
[Al
3+
]
2
γ
Al
2
(OH)
2

c
γ
H
2
H
+
[]
2

== =
K
Al
2
(OH)
2
c
K
sp,Al(OH)
3
2
γ
H
4
[H
+
]
4
γ
Al
2

(OH)
2
c
K
w
6

=
γ
Al
,
γ
H
,
γ
Al(OH)c
,
γ
Al
7
(OH)
17
c
γ
Al
13
(OH)
34
c
γ

Al(OH)
4
c
,
γ
Al
2
(OH)
2
c
Al
7
(OH)
17
4+
Al
13
(OH)
34
5+
Al(OH)
4
−
Al
2
(OH)
2
4+
K
sp

,Al(OH)
3
Al(OH)
3(s)
K
Al(OH)c
K
Al
7
(OH)
17
c
, K
Al
13
(OH)
34
c
K
Al(OH)
4
c
K
Al
2
(OH)
2
c
Al(OH)
2+

Al
7
(OH)
17
4+
Al
13
(OH)
34
5+
Al(OH)
4
−
,Al
2
(OH)
2
4+
∆H
298
o
TX249_frame_C12.fm Page 556 Friday, June 14, 2002 2:29 PM
© 2003 by A. P. Sincero and G. A. Sincero
Equations (12.12) through (12.17) may now be substituted into Equation (12.11)
to produce
(12.18)
To obtain the optimum pH, differentiate [sp
Al
] of Equation (12.18) with respect
to [H

+
] and equate the result to zero. Doing the differentiation, rearranging the
resulting equation, and calling the resulting solution for [ ] as [ ], obtain the
following equation:
(12.19)
By trial and error, Equation (15.19) may now be solved for [ ]. Thus, the
optimum pH is
(12.20)
Example 12.1 A raw water containing 140 mg/L of dissolved solids is sub-
jected to coagulation treatment using alum. Calculate the optimum pH that the
operation should be conducted. Assume the temperature of operation is 25°C.
Solution:
sp
Al
[]
K
sp,Al OH()
3
γ
H
3
[H
+
]
3
γ
Al
K
w
3

=
K
Al OH()c
K
sp,Al OH()
3
γ
H
2
[H
+
]
2
γ
Al(OH)c
K
w
3

+

7K
Al
7
(OH)
17
c
K
sp,Al OH()

3
7
γ
H
4
[H
+
]
4
γ
Al
7
(OH)
17
c
K
w
21

13K
Al
13
(OH)
34
c
K
sp,Al OH()
3
13
γ

H
5
[H
+
]
5
γ
Al
13
(OH)
34
c
K
w
39

++

K
Al(OH)
4
c
K
w
γ
Al(OH)
4
c
γ
H

[H
+
]

2K
Al
2
(OH)
2
c
K
sp,Al OH()
3
2
γ
H
4
[H
+
]
4
γ
Al
2
(OH)
2
c
K
w
6

++
H
+
H
opt
+
2K
Al(OH)c
K
sp,Al OH()
3
γ
H
2
γ
Al(OH)c
K
w
3




H
opt
+
[]
3
3K

sp,Al OH()
3
γ
H
3
γ
Al
K
w
3




H
opt
+
[]
4
+

28K
Al
7
(OH)
17
c
K
sp,Al OH()
3

7
γ
H
4
γ
Al
7
(OH)
17
c
K
w
21

8K
Al
2
(OH)
2
c
K
sp,Al OH()
3
2
γ
H
4
γ
Al
2

(OH)
2
c
K
w
6

+



H
opt
+
[]
5
+

65K
Al
13
(OH)
34
c
K
sp,Al OH()
3
13
γ
H

5
γ
Al
13
(OH)
34
c
K
w
39




H
opt
+
[]
6
+
K
Al(OH)
4
c
K
w
γ
Al(OH)
4
c

γ
H

=
H
opt
+
pH H
opt
+
{}log–
γ
H
[H
opt
+
]()log==
2K
Al(OH)c
K
sp,Al OH()
3
γ
H
2
γ
Al(OH)c
K
w
3




H
opt
+
[]
3
3K
sp,Al OH()
3
γ
H
3
γ
Al
K
w
3




H
opt
+
[]
4
+

28K
Al
7
(OH)
17
c
K
sp,Al OH()
3
7
γ
H
4
γ
Al
7
(OH)
17
c
K
w
21

8K
Al
2
(OH)
2
c

K
sp,Al OH()
3
2
γ
H
4
γ
Al
2
(OH)
2
c
K
w
6

+



H
opt
+
[]
5
+

65K
Al

13
(OH)
34
c
K
sp,Al OH()
3
13
γ
H
5
γ
Al
13
(OH)
34
c
K
w
39




H
opt
+
[]
6
+

K
Al(OH)
4
c
K
w
γ
Al(OH)
4
c
γ
H

=
TX249_frame_C12.fm Page 557 Friday, June 14, 2002 2:29 PM
© 2003 by A. P. Sincero and G. A. Sincero
Therefore,
Therefore,
Therefore,
2K
Al(OH)c
K
sp,Al OH()
3
γ
H
2
γ
Al(OH)c
K

w
3

:
K
Al(OH)c
10
−5
K
sp,Al OH()
3
10
−33
µ
2.5 10
−5
()TDS
γ
10
−
0.5z
i
2
(
µ
)
1+1.14
µ
()

=== =
µ
2.5 10
5–
()140()3.5 10
3–
()==
γ
H
10
−
0.5 1()
2
3.5 10
3–
()[]
1+1.14 3.5 10
3–
()[]

0.94==
γ
Al(OH)c
10
−
0.5 2()
2
3.5 10
3–
()[]

1+1.14 3.5 10
3–
()[]

0.77==K
w
10
14–
=
2K
Al(OH)c
K
sp,Al OH()
3
γ
H
2
γ
Al(OH)c
K
w
3

210
5–
()10
33–
()0.94()
2
0.77()10

14–
()
3

2.30 10
4
()==
3K
sp,Al OH()
3
γ
H
3
γ
Al
K
w
3

:
γ
Al
10
−
0.5 3()
2
3.5 10
3–
()[]
1+1.14 3.5 10

3–
()[]

0.56==
3K
sp,Al OH()
3
γ
H
3
γ
Al
K
w
3

310
33–
()0.94()
3
0.56()10
14–
()
3

4.45 10
9
()==
28K
Al

7
(OH)
17
c
K
sp,Al OH()
3
7
γ
H
4
γ
Al
7
(OH)
17
c
K
w
21

8K
Al
2
(OH)
2
c
K
sp,Al OH()
3

2
γ
H
4
γ
Al
2
(OH)
2
c
K
w
6

:+
K
Al
7
(OH)
17
c
10
48.8–
γ
Al
7
(OH)
17
c
γ

Al
2
(OH)
2
c
10
0.5 4()
2
3.5 10
3–
()[]
1+1.14 3.5 10
3–
()[]
–
0.36=== =
K
Al
2
(OH)
2
c
10
6.3–
=
28K
Al
7
(OH)
17

c
K
sp,Al OH()
3
7
γ
H
4
γ
Al
7
(OH)
17
c
K
w
21

8K
Al
2
(OH)
2
c
K
sp,Al OH()
3
2
γ
H

4
γ
Al
2
(OH)
2
c
K
w
6

+
28 10
48.8–
()10
33–
()
7
0.94()
4
0.36()10
14–
()
21

810
6.3–
()10
33–
()

2
0.94()
4
0.36()10
14–
()
6

+=
9.62 10
15
()= 8.70 10
12
()+ 9.63 10
15
()=
TX249_frame_C12.fm Page 558 Friday, June 14, 2002 2:29 PM
© 2003 by A. P. Sincero and G. A. Sincero
Therefore,
Therefore,
Therefore,
Solving by trial and error, let
[] Y
10
−6
4.66(10
−14
)
10
−8

2.30(10
−20
)
65K
Al
13
(OH)
34
c
K
sp,Al OH()
3
13
γ
H
5
γ
Al
13
(OH)
34
c
K
w
39

:
Κ
Al
13

(OH)
34
c
10
97.4–
=
γ
Al
13
(OH)
34
c
10
0.5 5()
2
3.5 10
3–
()[]
1+1.14 3.5 10
3–
()[]

–
0.20==
65K
Al
13
(OH)
34
c

K
sp,Al OH()
3
13
γ
H
5
γ
Al
13
(OH)
34
c
K
w
39

65 10
97.4–
()10
33–
()
13
0.94()
5
0.20()10
14–
()
39

65 0.94()
5
0.20()10
19.6–
()

==
9.50 10
21
()=
K
Al(OH)
4
c
K
w
γ
Al(OH)
4
c
γ
H

:
K
Al(OH)
4
c
10
+1.3

=
γ
Al(OH)
4
c
γ
H
0.94==
K
Al(OH)
4
c
K
w
γ
Al(OH)
4
c
γ
H

10
+1.3
()10
14–
()
0.94()0.94()

2.26 10
13–

()==
2.30 10
4
()H
opt
+
[]
+3
4.45 10
9
()H
opt
+
[]
4
9.63 10
15
()H
opt
+
[]
5
9.50 10
21
()H
opt
+
[]
6
+{++

2.26 10
13–
()=
Y 2.30 10
4
()H
opt
+
[]
3
= 4.45 10
9
()H
opt
+
[]
4
9.63 10
15
()H
opt
+
[]
5
9.50 10
21
()H
opt
+
[]

6
+{++
H
opt
++
++
y 2.26(10
13–
)
y 10
6–
–
10
6–
10
8–
–

2.26 10
13–
()4.66 10()
14–
–
4.66 10()
14–
2.30 10
20–
()–

=

10
6–
4.66(10
14–
) y 4.81 10
6–
()=
10
8–
2.30(10
20–
) Therefore, pH log
10
[4.81(10
6–
)]– 5.32 Ans==
TX249_frame_C12.fm Page 559 Friday, June 14, 2002 2:29 PM
© 2003 by A. P. Sincero and G. A. Sincero
12.6 CHEMICAL REACTIONS OF THE FERROUS ION
The ferrous salt used as coagulant in water and wastewater treatment is copperas,
FeSO
4
⋅ 7H
2
O. For brevity, this will simply be written without the water of hydration
as FeSO
4
. When copperas dissolves in water, it dissociates according to the following
equation:
(12.21)

As in the case of alum, the ions must be rapidly dispersed throughout the tank
in order to effect the complete coagulation process. The solid precipitate Fe(OH)
2(s)
and complexes are formed and expressing in terms of equilibrium with the solid
Fe(OH)
2(s)
, the following reactions transpire (Snoeyink and Jenkins, 1980):
(12.22)
(12.23)
(12.24)
The complexes are FeOH
+
and . Also note that the OH
−
ion is a
participant in these reactions. This means that the concentrations of each of these
complex ions are determined by the pH of the solution. In the application of the
above equations in an actual coagulation treatment of water, conditions must be
adjusted to allow maximum precipitation of the solid represented by Fe(OH)
2(s)
. To
allow for this maximum precipitation, the concentrations of the complex ions must
be held to the minimum. The values of the equilibrium constants given above are
at 25°C.
12.6.1 DETERMINATION OF THE OPTIMUM pH
For effective removal of the colloids, as much of the copperas should be converted
to the solid Fe(OH)
2(s)
. Also, as much of the concentrations of the complex ions
should neutralize the primary charges of the colloids to effect their destabilization.

Overall, this means that once the solids have been formed and the complex ions
have neutralized the colloid charges, the concentrations of the complex ions standing
in solution should be at the minimum. The pH corresponding to this condition is
the optimum pH for the coagulation using copperas.
Let sp
FeII
represent all the species that contain the Fe(II) ion standing in solution.
Thus, the concentration of all the species containing the ion is
(12.25)
All the concentrations in the right-hand side of the previous equation will now be
expressed in terms of the hydrogen ion concentration. As in the case of alum, this
will result in the expressing of [sp
FeII
] in terms of the hydrogen ion. Differentiating
the resulting equation of [sp
FeII
] with respect to [H
+
] and equating the result to zero
FeSO
4
Fe
2+
→ SO
4
2−
+
Fe(OH)
2 s()
 Fe

2+
2OH
−
+ K
sp,Fe OH()
2
10
14.5–
=
Fe(OH)
2 s()
 FeOH
+
OH
−
+ K
FeOHc
10
9.4–
=
Fe(OH)
2 s()
OH
−
 Fe OH()
3
−
+ K
Fe OH()
3

c
10
5.1–
=
Fe(OH)
3
−
sp
FeII
[]Fe
2+
[]= FeOH
+
[]Fe(OH)
3
−
[]++
TX249_frame_C12.fm Page 560 Friday, June 14, 2002 2:29 PM
© 2003 by A. P. Sincero and G. A. Sincero
will produce the minimum concentration of sp
FeII
and, thus, the optimum pH deter-
mined. Using the equilibrium reactions, Eqs. (12.22) through (12.24), along with
the ion product of water, we now proceed as follows:
(12.26)
(12.27)
(12.28)
γ
FeII
,

γ
FeOHc
, are, respectively, the activity coefficients of the ferrous ion
and the complexes FeOH
+
and [ ]. is the solubility product constant
of the solid Fe(OH)
2(s)
. K
FeOHc
and are, respectively, the equilibrium con-
stants of the complexes FeOH
+
and .
Equations (12.26) through (12.28) may now be substituted into Equation (12.25)
to produce
(12.29)
Differentiating with respect to [H
+
], equating to zero, rearranging, and changing H
+
to , the concentration of the hydrogen ion at optimum conditions,
(12.30)
The value of [H
opt
] may be solved by trial error.
12.7 CHEMICAL REACTIONS OF THE FERRIC ION
The ferric salts used as coagulant in water and wastewater treatment are FeCl
3
and

Fe
2
(SO
4
)
3
. They have essentially the same chemical reactions in that both form the
Fe(OH)
3(s)
solid. When these coagulants are dissolved in water, they dissociate
according to the following equations:
(12.31)
(12.32)
Fe
2+
[]
Fe
2+
{}
γ
FeII

K
sp,Fe(OH)
2
γ
FeII
OH
−
{}

2

K
sp,Fe(OH)
2
{H
+
}
2
γ
FeII
K
w
2

K
sp,Fe(OH)
2
γ
H
2
[H
+
]
2
γ
FeII
K
w
2

== = =
FeOH
+
[]
FeOH
+
{}
γ
FeOHc

K
FeOHc
γ
FeOHc
{OH
−
}

K
FeOHc
{H
+
}
γ
FeOHc
K
w

K

FeOHc
γ
H
[H
+
]
γ
FeOHc
K
w

== = =
Fe(OH)
3
−
[]
Fe(OH)
3
−
{}
γ
Fe(OH)
3
c

K
Fe(OH)
3
c
{OH

−
}
γ
Fe(OH)
3
c

K
Fe(OH)
3
c
K
w
γ
Fe(OH)
3
c
{H
+
}

K
Fe(OH)
3
c
K
w
γ
Fe(OH)
3

c
γ
H
[H
+
]

== = =
γ
Fe(OH)
3
c
Fe(OH)
3
−
K
sp,Fe(OH)
2
K
Fe(OH)
3
c
Fe(OH)
3
−
sp
FeII
[]
K
sp,Fe(OH)

2
γ
H
2
[H
+
]
2
γ
FeII
K
w
2

=
K
FeOHc
γ
H
[H
+
]
γ
FeOHc
K
w

K
Fe(OH)
3

c
K
w
γ
Fe(OH)
3
c
γ
H
[H
+
]

++
H
opt
+
2K
sp,Fe(OH)
2
γ
H
2
γ
FeII
K
w
2





H
opt
+
[]
3
K
FeOHc
γ
H
γ
FeOHc
K
w




H
opt
+
[]
2
+
K
Fe(OH)
3
c
K

w
γ
Fe(OH)
3
c
γ
H

=
FeCl
3
Fe
3+
3Cl
−
+→
Fe
2
(SO
4
)
3
2Fe
3+
3SO
4
2−
+→
TX249_frame_C12.fm Page 561 Friday, June 14, 2002 2:29 PM
© 2003 by A. P. Sincero and G. A. Sincero

As in any coagulation process, these ions must be rapidly dispersed throughout the
tank in order to effect the complete coagulation process. The solid precipitate
Fe(OH)
3(s)
and complexes are then formed. The reactions, together with the respective
equilibrium constants at 25°C, are as follows (Snoeyink and Jenkins, 1980):
(12.33)
(12.34)
(12.35)
(12.36)
(12.37)
The complexes are FeOH
2+
, , , and . Also note that
the OH
−
ion is a participant in these reactions. This means that the concentrations
of each of these complex ions are determined by the pH of the solution.
In the application of the above equations in an actual coagulation treatment of
water as in all applications of coagulants, conditions must be adjusted to allow
maximum precipitation of the solid which in the present case is represented by
Fe(OH)
3(s)
. To allow for this maximum precipitation, the concentrations of the
complex ions must be held to a minimum.
12.7.1 DETERMINATION OF THE OPTIMUM pH
For effective removal of the colloids, as much of the ferric ions should be converted
to the solid Fe(OH)
3(s)
. Also, as much of the concentrations of the complex ions

should neutralize the primary charges of the colloids to effect their destabilization.
Overall, this means that once the solids have been formed and the complex ions
have neutralized the colloid charges, the concentrations of the complex ions standing
in solution should be at the minimum, which corresponds to the optimum pH for
the coagulation process.
Let sp
FeIII
represent all the species that contain the Fe(III) ion standing in solution.
Thus, the concentration of all the species containing the ion is
(12.38)
All the concentrations in the right-hand side of the above equation will now be
expressed in terms of the hydrogen ion concentration. This will result in the expressing
of [sp
FeIII
] in terms of the hydrogen ion. Differentiating the resulting equation of
Fe OH()
3 s()
 Fe
3+
3OH
−
+ K
sp Fe OH()
3
,
10
38–
=
Fe OH()
3 s()

 FeOH
2+
2OH
−
+ K
FeOHc
10
26.16–
=
Fe OH()
3 s()
 Fe OH()
2
+
OH
−
+ K
Fe OH()
2
c
10
16.74–
=
Fe OH()
3 s()
OH
−
 Fe OH()
4
−

+ K
Fe OH()
4
c
10
5–
=
2Fe OH()
3 s()
 Fe
2
OH()
2
4+
4OH
−
+ K
Fe
2
OH()
2
c
10
50.8–
=
Fe(OH)
2
+
Fe(OH)
4

−
Fe
2
(OH)
2
4+
sp
FeIII
[]Fe
3+
[]= FeOH
2+
[]Fe(OH)
2
+
[]Fe(OH)
4
−
[]2Fe
2
(OH)
2
4+
[]++ + +
TX249_frame_C12.fm Page 562 Friday, June 14, 2002 2:29 PM
© 2003 by A. P. Sincero and G. A. Sincero
[sp
FeIII
] with respect to [H
+

] and equating the result to zero will produce the minimum
concentration of sp
FeIII
and, thus, the optimum pH determined. Using the equilibrium
reactions, Eqs. (12.33) through (12.37), along with the ion product of water, we now
proceed as follows:
(12.39)
(12.40)
(12.41)
(12.42)
(12.43)
γ
FeIII
,
γ
H
,
γ
FeOHc
, are, respectively, the activity coeffi-
cients of the ferric and the hydrogen ions and the complexes FeOH
2+
, ,
and is the solubility product constant of the solid
Fe(OH)
3(s)
and K
w
is the ion product of water. K
FeOHc

, and
are, respectively, the equilibrium constants of the complexes FeOH
2+
,
, , and .
Equations (12.39) through (12.43) may now be substituted into Equation (12.38)
to produce
(12.44)
Fe
3+
[]
Fe
3+
{}
γ
FeIII

K
sp,Fe(OH)
3
γ
FeIII
{OH
−
}
3

K
sp,Fe(OH)
3

{H
+
}
3
γ
FeIII
K
w
3

K
sp,Fe(OH)
3
γ
H
3
[H
+
]
3
γ
FeIII
K
w
3

== = =
FeOH
2+
[]

FeOH
2+
{}
γ
FeOHc

K
FeOHc
γ
FeOHc
{OH
−
}
2

K
FeOHc
{H
+
}
2
γ
FeOHc
K
w
2

K
FeOHc
γ

H
2
[H
+
]
2
γ
FeOHc
K
w
2

== = =
Fe(OH)
2
+
[]
Fe(OH)
2
+
{}
γ
Fe(OH)
2
c

K
Fe(OH)
2
c

γ
Fe(OH)
2
c
{OH
−
}

K
Fe(OH)
2
c
{H
+
}
γ
Fe(OH)
2
c
K
w

K
Fe(OH)
2
c
γ
H
[H
+

]
γ
Fe(OH)
2
c
K
w

== = =
Fe(OH)
4
−
[]
Fe(OH)
4
−
{}
γ
Fe(OH)
4
c

K
Fe(OH)
4
c
{OH
−
}
γ

Fe(OH)
4
c

K
Fe(OH)
4
c
K
w
γ
Fe(OH)
4
c
{H
+
}

K
Fe(OH)
4
c
K
w
γ
Fe(OH)
4
c
γ
H

{H
+
}

== = =
Fe
2
(OH)
2
4+
[]
Fe
2
(OH)
2
4+
{}
γ
Fe
2
(OH)
2
c

K
Fe
2
(OH)
2
c

γ
Fe
2
(OH)
2
c
{OH
−
}
4

K
Fe
2
(OH)
2
c
γ
H
4
[H
+
]
4
γ
Fe
2
(OH)
2
c

K
w
4

== =
γ
Fe(OH)
2
c
,
γ
Fe(OH)
4
c
,
γ
Fe
2
(OH)
2
c
Fe(OH)
2
+
Fe(OH)
4
−
,Fe
2
(OH)

2
4+
. K
sp, Fe(OH)
3
K
Fe(OH)
2
c
,
K
Fe(OH)
4
c
,
K
Fe
2
(OH)
2
c
Fe(OH)
2
+
Fe(OH)
4
−
Fe
2
(OH)

2
4+
[sp
FeIII
]
K
sp,Fe(OH)
3
γ
H
3
[H
+
]
3
γ
FeIII
K
w
3

=
K
FeOHc
γ
H
2
[H
+
]

2
γ
FeOHc
K
w
2

K
Fe(OH)
2
c
γ
H
[H
+
]
γ
Fe(OH)
2
c
K
w

++
K
Fe(OH)
4
c
K
w

γ
Fe(OH )
4
c
{H
+
}

2K
Fe
2
(OH)
2
c
γ
H
4
[H
+
]
4
γ
Fe
2
(OH)
2
c
K
w
4

++
TX249_frame_C12.fm Page 563 Friday, June 14, 2002 2:29 PM
© 2003 by A. P. Sincero and G. A. Sincero
Differentiating with respect to [H
+
], equating to zero, rearranging, and changing H
+
to , the concentration of the hydrogen ion at optimum conditions,
(12.45)
The value of [H
opt
] may be solved by trial error.
Example 12.2 A raw water containing 140 mg/L of dissolved solids is sub-
jected to coagulation treatment using copperas. Calculate the optimum pH that the
operation should be conducted. Assume the temperature of operation is 25°C.
Solution:
Therefore,
Therefore,
H
opt
+
8K
Fe
2
(OH)
2
c
γ
H

4
γ
Fe
2
(OH)
2
c
K
w
4




H
opt
+
[]
5
3K
sp,Fe(OH)
3
γ
H
3
γ
FeIII
K
w
3




H
opt
+
[]
4
2K
FeOHc
γ
H
2
γ
FeOHc
K
w
2




H
opt
+
[]
3
++
K

Fe(OH)
2
c
γ
H
γ
Fe(OH)
2
c
K
w




H
opt
+
[]
2
+
K
Fe(OH)
4
c
K
w
γ
Fe(OH)
4

c
γ
H

=
2K
sp,Fe(OH)
2
γ
H
2
γ
FeII
K
w
2




H
opt
+
[]
3
K
FeOHc
γ
H
γ

FeOHc
K
w




H
opt
+
[]
2
+
K
Fe(OH)
3
c
K
w
γ
Fe(OH)
3
c
γ
H

=
2K
sp,Fe(OH)
2

γ
H
2
γ
FeII
K
w
2

:
K
sp,Fe(OH)
2
10
14.5–
=
γ
H
0.94=
γ
FeII
10
−
0.5 2()
2
3.5 10
3–
()[]
1+1.14 3.5 10
3–

()[]

0.77==
2K
sp,Fe(OH)
2
γ
H
2
γ
FeII
K
w
2

210
14.5–
()0.94()
2
0.77()10
14–
()
2

7.26 10
13
()==
K
FeOHc
γ

H
γ
FeOHc
K
w

:
K
FeOHc
10
9.4–
=
γ
FeOHc
γ
H
0.94==
K
FeOHc
γ
H
γ
FeOHc
K
w

10
9.4–
()0.94()
0.94()10

14–
()

3.98 10
4
()==
K
Fe(OH)
3
c
K
w
γ
Fe(OH)
3
c
γ
H

:
K
Fe(OH)
3
c
10
5.1–
=
γ
Fe(OH)
3

c
γ
H
0.94==
K
Fe(OH)
3
c
K
w
γ
Fe(OH)
3
c
γ
H

10
5.1–
()10
14–
()
0.94()0.94()

8.99 10
20–
()==
TX249_frame_C12.fm Page 564 Friday, June 14, 2002 2:29 PM
© 2003 by A. P. Sincero and G. A. Sincero
Therefore,

Solving by trial and error, let
Example 12.3 A raw water containing 140 mg/L of dissolved solids is sub-
jected to coagulation treatment using a ferric salt. Calculate the optimum pH that
the operation should be conducted. Assume the temperature of operation is 25°C.
Solution:
Therefore,
[] Y
10
−11
4.05(10
−18
)
10
−12
3.99(10
−20
)
7.26(10
13
)H
opt
+
[]
3
3.98 10
4
()H
opt
+
[]+ 8.99 10

20–
()=
Y 7.26 10
13
()H
opt
+
[]
3
= 3.98 10
4
()H
opt
+
[]
2
+
H
opt
++
++
10
11–
4.05 10
18–
()
y 10
12–
–
10

12–
10
11–
–

8.99 10
20–
()3.99 10
20–
()–
3.99 10
20–
()4.05 10
18–
()–

=
y 8.99 10
20–
()y 1.11(10
12–
)=
10
12–
8.99 10
20–
()Therefore, pH log
10
1.11 10
12–

()[]– 11.95 Ans==
8K
Fe
2
(OH)
2
c
γ
H
4
γ
Fe
2
(OH)
2
c
K
w
4




H
opt
+
[]
5
3K
sp,Fe(OH)

3
γ
H
3
γ
FeIII
K
w
3




H
opt
+
[]
4
2K
FeOHc
γ
H
2
γ
FeOHc
K
w
2





H
opt
+
[]
3
++
+
K
Fe(OH)
2
c
γ
H
γ
Fe(OH)
2
c
K
w




H
opt
+
[]
2

K
Fe(OH)
4
c
K
w
γ
Fe(OH)
4
c
γ
H

=
8K
Fe
2
(OH)
2
c
γ
H
4
γ
Fe
2
(OH)
2
c
K

w
4

:
K
Fe
2
(OH)
2
c
10
50.8–
=
γ
H
0.94=
γ
Fe
2
(OH)
2
c
10
0.5(4)
2
3.5 10
3–
()[]
1+1.14 3.5 10
3–

()[]
–
0.36==
8K
Fe
2
(OH)
2
c
γ
H
4
γ
Fe
2
(OH)
2
c
K
w
4

8(10
50.8–
)(0.94)
4
(0.36)(10
14–
)
4

2.75(10
6
)==
3K
sp,Fe(OH)
3
γ
H
3
γ
FeIII
K
w
3

:
K
sp,Fe(OH)
3
10
38–
=
γ
FeIII
10
0.5(3)
2
3.5(10
3–

)[]
1+1.14 3.5 10
3–
()[]
–
0.56==
TX249_frame_C12.fm Page 565 Friday, June 14, 2002 2:29 PM
© 2003 by A. P. Sincero and G. A. Sincero
Therefore,
Therefore,
Therefore,
Therefore,
Therefore,
Solving by trial and error, let
3K
sp,Fe(OH)
3
γ
H
3
γ
FeIII
K
w
2

3(10
38–
)(0.94)
3

0.56(10
14–
)
3

4.45(10
4
)==
2K
FeOHc
γ
H
2
γ
FeOHc
K
w
2

:
K
FeOHc
10
26.16–
=
γ
FeOHc
10
0.5(2)
2

3.5(10
3–
)[]
1+1.14 3.5(10
3–
)[]
–
0.77==
2K
FeOHc
γ
H
2
γ
FeOHc
K
w
2

2(10
26.16–
)(0.94)
2
(0.77)(10
14–
)
2

158.78==
K

Fe(OH)
2
c
γ
H
γ
Fe(OH)
2
c
K
w

:
K
Fe(OH)
2
c
10
16.74–
=
γ
Fe(OH)
2
c
γ
H
0.94==
K
Fe(OH )
2

c
γ
H
γ
Fe(OH)
2
c
K
w

10
16.74–
(0.94)
(0.94)(10
14–
)

10
2.74–
==
K
Fe(OH)
4
c
K
w
γ
Fe(OH)
4
c

γ
H

:
K
Fe(OH)
4
c
10
5–
=
γ
Fe(OH)
4
c
γ
H
0.94==
K
Fe(OH )
4
c
K
w
γ
Fe(OH)
4
c
γ
H

10
5–
(10
14–
)
0.94(0.94)

1.13(10
19–
)==
2.75(10
6
)[H
opt
+
]
5
4.45(10
4
)[H
opt
+
]
4
158.78[H
opt
+
]
3

10
2.74–
[H
opt
+
]
2
+++1.13(10
19–
)=
5(12) 60= 4(12) 48=
Y 2.75(10
6
)[H
opt
+
]
5
= 4.45(10
4
)[H
opt
+
]
4
158.78[H
opt
+
]
3

10
2.74–
[H
opt
+
]
2
+++
TX249_frame_C12.fm Page 566 Friday, June 14, 2002 2:29 PM
© 2003 by A. P. Sincero and G. A. Sincero
12.8 JAR TESTS FOR OPTIMUM pH DETERMINATION
We may summarize the optimum pH’s of the coagulants obtained in the previous
examples: alum = 5.32, ferrous = 11.95, and ferric = 8.2. The problem with these
values is that they only apply at a temperature of 25°C. If the formulas for the
determination of these pH’s are reviewed, they will be found to be functions of
equilibrium constants. By the use of the Van’t Hoff equation, values at other tem-
peratures for the equilibrium constants can be found. These, however, as mentioned
before, also need the value of the standard enthalpy change, as discussed in
the chapter on water stabilization. For the aforementioned coagulants, no values of
the enthalpy change are available. Thus, until studies are done to determine these
values, optimum pH values must be determined using the jar test.
In addition, the optimum pH’s of 5.32, 11.95, and 8.2 were obtained at a dissolved
solids of 140 mg/L. The value of the dissolved solids predicts the values of the activity
coefficients of the various ions in solution, which, in turn, determine the activities of
the ions, including that of the hydrogen ion. It follows that, if the dissolved solids
concentration is varied, other values of optimum pH’s will also be obtained not only
the respective values of 5.32, 11.95, and 8.2. This is worth repeating: the values of
5.32, 11.95, and 8.2 apply only at a dissolved solids concentration of 140 mg/L. In
addition, they only apply provided the temperature is 25°C. In subsequent discussions,
mention of these optimum pH values would mean values at the conditions of 25°C of

temperature and a solids concentration of 140 mg/L.
12.9 CHEMICAL REQUIREMENTS
The chemical reactions written so far are not usable for determining chemical
requirements. They all apply at equilibrium conditions. When chemicals are added
to water, it first has to react, after which equilibrium will set in. It is this first reaction
that determines the amount of chemical, not when equilibrium has set in. Chemical
requirements for the three methods of coagulation treatments discussed will now be
addressed.
The respective chemical reactions will first be derived. From the concept of
equivalence, the number of equivalents of all the species participating in a given
[] Y
10
−8
1.82(10
−19
)
10
−10
1.82(10
−23
)
H
opt
++
++
10
8–
1.82 10
19–
()

y 10
8–
–
10
8–
10
10–
–

1.13(10
19–
) 1.82(10
19–
)–
1.82(10
19–
) 1.82(10
23–
)–

=
y 1.13(10
19–
) y 0.63(10
8–
)=
10
10–
1.82 10
23–

()Therefore, pH log
10
[0.63(10
8–
)]– 8.2 Ans==
∆H
298
o
,
TX249_frame_C12.fm Page 567 Friday, June 14, 2002 2:29 PM
© 2003 by A. P. Sincero and G. A. Sincero
chemical reaction are equal. From the reaction, the equivalent masses will then be
calculated. Once this is done, equations for chemical requirements can be derived.
12.9.1 CHEMICAL REQUIREMENTS IN ALUM
C
OAGULATION TREATMENT
The water of hydration for alum varies from 13 to 18. For the purposes of calculating
the chemical requirements, this range of values will be designated by x. In actual
applications, the correct value of x must be obtained from the label of the container
used to ship the chemical. Using x as the water of hydration, the chemical reaction
for alum is

(12.46)
The rationale behind the previous reaction is explained here. The bicarbonate is
known to act as a base as well as an acid. As a base, its interaction is
. The K
sp
of aluminum hydroxide is approx-
imately (10
−33

). This means that the hydroxide is very insoluble. Thus, with the ions
of the coagulant and the bicarbonate dispersed in the water, Al
3+
‘‘grabs” whatever
OH
−
there is to form the precipitate, Al(OH)
3
, and the reaction portrayed above ensues.
A very important point must be discussed with respect to the previous coagulation
reaction, in comparison with those found in the literature. The environmental engi-
neering literature normally uses the equilibrium arrows,  , instead of the single
forward arrow, → , as written previously. Equilibrium arrows indicate that a particular
reaction is in equilibrium, which would mean for the present case, that alum is produced
in the backward reaction. Alum, however, is never produced by mixing aluminum
hydroxide, carbon dioxide, calcium sulfate, and water, the species found on the right-
hand side of the previous equation. Once Al
2
(SO
4
)
3
⋅ 14H
2
O is mixed with Ca(HCO
3
)
2
,
the alum is gone forever producing the aluminum hydroxide precipitate—it cannot be

recovered. After the formation of the precipitate, any backward reaction would be for
the complex reactions and not for the formation of Al
2
(SO
4
)
3
⋅ xH
2
O, as would be
inferred if the above reaction were written with the equilibrium arrows.
In addition, coagulation is a process of expending the coagulant. In the process
of expenditure, the alum must react to produce its products. This means that what
must ‘‘exist” is the forward arrow and not any backward arrow. Portraying the
backward arrow would mean that the alum is produced, but it is known that it is not
produced but expended. During expenditure, no equilibrium must exist. To reiterate,
the coagulation reaction should be represented by the forward arrow and not by the
equilibrium arrows.
As shown in Equation (12.46), an alkaline substance is needed to react with the
alum. The bicarbonate alkalinity is used, since it is the alkalinity that is always found
in natural waters. In practice, its concentration must be determined to ascertain if
enough is present to satisfy the optimum alum dose. If found deficient, then lime is
normally added to satisfy the additional alkalinity requirement. As we have found,
the reaction is optimum at a pH of 5.32 at 25°C when the dissolved solids concen-
tration is 140 mg/L.
Al
2
(SO
4
)

3
xH
2
O⋅ 3Ca(HCO
3
)
2
+ 2Al(OH)
3
↓
→ 6CO
2
3CaSO
4
xH
2
O++ +
HCO
3
−
H
2
O+
 H
2
CO
3
OH
−
+  CO

2
H
2
OOH
−
++
TX249_frame_C12.fm Page 568 Friday, June 14, 2002 2:29 PM
© 2003 by A. P. Sincero and G. A. Sincero
In any given application, it is uncertain whether the bicarbonate alone or in
combination is needed, therefore, the most practical way of expressing the alkalinity
requirement is through the use of equivalents. Using this method, the number of
equivalents of the alum used is equal to the number of equivalents of the alkalinity
required; in fact, it is equal to the same number of equivalents of any species
participating in the chemical reaction. All that is needed, therefore, is to find the
number of equivalent masses of the alum and the alkalinity needed is equal to this
number of equivalent masses.
In the previous reaction, the number of references is 6. Thus, the equivalent
mass of alum is Al
2
(SO
4
)
3
⋅ xH
2
O/6 = 57.05 + 3x and that of the calcium bicarbonate
species is 3Ca(HCO
3
)
2

/6 = 81.05. The other alkalinity sources that can be used are
lime, caustic soda, and soda ash. Lime is used in the discussion that follows. Also,
alkalinity requirements are usually expressed in terms of CaCO
3
. Therefore, we also
express the reactions of alum in terms of calcium carbonate. The respective chemical
reactions are:
(12.47)
(12.48)
Ca(OH)
2
is actually slaked lime: CaO + H
2
O. Note that in order to find the equivalent
masses, the same number of molecules of alum in the balanced chemical reaction
as used in Equation (12.46) should be used in Eqs. (12.47) and (12.48); otherwise,
the equivalent masses obtained are equivalent to each other. From the reactions, the
equivalent mass of lime (CaO) is and that of calcium carbonate
is .
It is impossible to determine the optimum dose of alum using chemical reaction.
This value must be obtained through the jar test. Let [Alopt]
mg
and [Alopt]
geq
be the
milligrams per liter and gram equivalents per liter of optimum alum dose, respec-
tively and let be the cubic meters of water or wastewater treated. Also, let M
CaOkgeqAl
and M
CaOAl

be the kilogram equivalents and kilogram mass of lime, respectively,
used at a fractional purity of P
CaO
. In the case of Ca(HCO
3
)
2
, the respective symbols
are and at a fractional purity of . = 1
in natural waters. Note the Al is one of the subscripts. This is to differentiate when
ferrous and ferric are used as the coagulants.
The number of kilogram equivalents of alum needed is ([Alopt]
geq
(1000)/1000)
= [Alopt]
geq
= ([Alopt]
mg
/1000(57.05 + 3x)) . Let M
Alkgeq
and M
Al
be the kilogram
equivalents and kilograms of alum used, respectively, at a fractional purity of P
Al
.
Thus,
(12.49)
(12.50)
Al

2
(SO
4
)
3
xH
2
O⋅ 3Ca(OH)
2
+ 2Al(OH)
3
→
↓
3CaSO
4
xH
2
O++
Al
2
(SO
4
)
3
xH
2
O⋅ 3CaCO
3
3HOH++ 2Al(OH)
3

↓
→ 3CaSO
4
3CO
2
xH
2
O+++
3CaO/6 28.05=
3CaCO
3
/6 50=
V
M
Ca(HCO
3
)
2
kgeqAl
M
Ca(HCO
3
)
2
Al
P
Ca(HCO
3
)
2

P
Ca(HCO
3
)
2
V V V
M
Alkgeq
Alopt[]
mg
1000(57.05 3x)P
Al
+

=
V

M
Al
Alopt[]
mg
1000P
Al

=
V
TX249_frame_C12.fm Page 569 Friday, June 14, 2002 2:29 PM
© 2003 by A. P. Sincero and G. A. Sincero